Let $A$ be a abelian group generated by elements $\langle a_1,a_2,a_3\rangle$ and $B$ be a subgroup generated by $\langle b_1,b_2,b_3\rangle$ where $\begin{pmatrix} b_1\\ b_2\\b_3 \end{pmatrix}= \begin{pmatrix} a_1&a_2&0\\ a_1&0&a_3\\ 0&a_2&a_3 \end{pmatrix}\begin{pmatrix}\alpha\\ \beta\\ \gamma\end{pmatrix}$ and $\alpha,\beta, \gamma \in \mathbb Z$

How then might we express $A/B$ as $\bigoplus_i \mathbb Z_{m_i}$?

  • $\begingroup$ sierra.nmsu.edu/morandi/notes/SmithNormalForm.pdf $\endgroup$ – Dylan Moreland Mar 17 '12 at 18:18
  • $\begingroup$ @DylanMoreland: Is the smith form always achievable? I am not so sure... $\endgroup$ – alex b Mar 17 '12 at 20:39
  • $\begingroup$ @DylanMoreland: Say $\alpha=5,\beta=3,\gamma=10$? $\endgroup$ – alex b Mar 17 '12 at 21:04
  • $\begingroup$ @alexb: Yes, the Smith Normal Form is always achievable. $\endgroup$ – Arturo Magidin Mar 17 '12 at 21:11
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    $\begingroup$ @Arturo, I hope you find that 5 that you lost in your head - you might need it some day. $\endgroup$ – Gerry Myerson Mar 19 '12 at 5:53

You should find the Smith decomposition of the given matrix, which gives you exactly the needed order for the sequence of cyclic groups. A reference for this process: Algebra, by Michael Artin.

  • $\begingroup$ Thank you, Rolando, but I am not sure tht the SNF is always achievable? Say $\alpha=5,\beta=3,\gamma=10$? $\endgroup$ – alex b Mar 17 '12 at 21:06
  • $\begingroup$ Please ignore the comment above, I can't seem to delete it. $\endgroup$ – alex b Mar 17 '12 at 21:19

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